[tex]y= \frac{log_3(x)}{2}[/tex]
There is a general formula that
[tex]log_a(x)= \frac{\log_b(x)}{log_b(a)}[/tex]
where b is any positive number. In particular, taking b= 3 here
[tex]log_9(x)= \frac{\log_3(x)}{log_3(9)}= \frac{\log_3(x)}{2}[/tex]

Now i have

log_3(x)
log_3(x) - -------- = 2
2

log3(x)- (1/2)log3(x)= 2

but i am really stuck with dealing with the x please help!!!!!

Could you solve A- (1/2)A= 2? If so then set log3(x) equal to that and solve for x by taking 3 to the power of each side.